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3 years ago

Calculate the RMS voltage of the following waveforms with 10 V peak-to-peak:

Physics

1 answer:

Deffense [45]3 years ago

6 0

Answer:

a) $T=0.01s$

b) $T=0.001s$

c) $T=0.00001s$

Explanation:

From the question we are told that:

Given Frequencies

a. 100 Hz,

b. 1 kHz,

c. 100 kHz.

Generally the equation for Waveform Period is mathematically given by

$T=\frac{1}{f}$

Therefore

For

$T=100 Hz$

$T=\frac{1}{100}$

$T=0.01s$

For

$F=1kHz$

$T=\frac{1}{1000}$

$T=0.001s$

For

$F=100kHz$

$T=\frac{1}{100*100}$

$T=0.00001s$

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Find the equivalent resistance, current, and voltage across each resistor when the specified resistors are connected across a 20

timama [110]

Answer:

Explanation:

The question is incomplete. Here is the complete question.

"Find the equivalent resistance, the current supplied by the battery and the current through each resistor when the specified resistors are connected across a 20-V battery. Part (a) uses two resistors with resistance values that can be set with the animation sliders, and you can use the animation to verify your calculation. In part (b), three resistors are specified. (a) Two resistors are connected in series across a 20-V battery. Let R1 = 1 Ω and R2 = 2 Ω. Rea = (b) Add a third resistor to the circuit in series. Let R1 = 1 Ω, R2 = 2 Ω, and R3 = 3 Ω"

Using ohms law formula to solve the problem

E = IRt

E is the supply voltage

I is the total current

Rt is the total equivalent resistant.

a) Given two resistances

R1 = 1ohms and R2 = 2ohms

If the resistors are Connected in series across a 20V supply voltage,

-Equivalent resistance = R1+R2

= 1ohms + 2ohms

= 3ohms

- In a series connected circuit, same current flows through the resistors.

Using the formula E = IRt

I = E/Rt

I = 20/3

I = 6.67A

The current in both resistors is 6.67A

- Different voltage flows across a series connected circuit.

Using the formula V = IR

V is the voltage across each resistor

I is the current in each resistor

For 1ohms resistor,

V = 6.67×1

V = 6.67Volts

For 2ohms resistor

V = 6.67×2

V = 13.34Volts

b) If the resistors are three

R1 = 1ohms, R2 = 2ohms R3 = 3ohms

- Total equivalent resistance = 1+2+3

= 6ohms

- Current in each resistor I = E/Rt

I = 20/6

I = 3.33A

Since the same current flows through the resistors, the current across each of them is 3.33A

- Voltage across them is calculated as shown:

V = IR

For 1ohm resistor

V = 3.33×1

V = 3.33volts

For 2ohms resistor

V = 3.33×2

V = 6.66volts

For 3ohms resistor

V = 3.33×3

V = 9.99volts

3 0

3 years ago

Read 2 more answers

NASA is designing a Mars-lander that will enter the Martian atmosphere at high speed. To land safely it must slow to a constant

Viktor [21]

Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

c) area of the parachute should be 12.68 m²

Explanation:

Given the data in the question;

V = 20 m/s

A = 200 m²

drag co-efficient CD = 1.855

g = 3.71 m/s²

density of the atmospheric pressure β = 0.01 kg/m³

a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²

FD = 742 N

we know that;

FD = Fg

Fg = gravity force

Fg = mg

FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;

Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

FD = 1780.8 N

Now, FD = 1/2 × CD × β × A × V², we solve for A

A = FD / 0.5 × CD × β × V²

we substitute

A = 1780.8 / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

A = 480 m²

Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

g = 9.8 m/s²,

β" = 1 kg/m³

v" = 20 m/s

M" = 480 kg

we know that;

FD = Fg = M"g

FD = 480 kg × 9.8 m/s² = 4704 N

from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

3 0

2 years ago

Can someone check if what i wrote so far makes sense and if i made a mistake.

Serjik [45]

I don't really know what it's about but everything looks okay to me. There might be some mistakes on the last sentence but i'm not completely sure.

7 0

3 years ago

Read 2 more answers

Suppose that you make a series RC circuit with a capacitor and a known resistor that has a 5% tolerance: R= 5.20 ± 0.26kΩ. You p

Masja [62]

Answer:

correct answer is C

Explanation:

The time constant of an RC circuit is

τ = RC

so to find the capacitance

C = τ/ R

C = 2.150 / 5.20 10³

C = 4.13 10⁻⁴ F

to find the error we use the worst case

ΔC = | $|\frac{dC}{d \tau }| \ \Delta \tau + | \frac{dC}{dR} | \ \Delta R$

the absolute value guarantees that we find the worst case, we evaluate the derivatives

ΔC = 1 /R Δτ + τ/R² ΔR

the absolute values of the errors are

Δτ = 0.002 s

ΔR = 0.3 kΩ

we substitute

ΔC = 0.002 /5.20 10³ + 2.150/(5.20 10³)² 0.3 10³

ΔC = 3.8 10⁻⁷ + 1.74 10⁻⁵

ΔC = 1.77 10⁻⁵ F

the uncertainty or error must be expressed with a significant figure

ΔC = 2 10⁻⁵ F

the percentage error is

Er% = $\frac{\Delta C}{C} \ 100$

Er% = $\frac{2 \ 10^{-5} }{ 4.13 \ 10^{-4} } \ 100$

Er% = 4.8%

the correct answer is C

3 0

3 years ago

An ultrasound pulse used in medical imaging has a frequency of 5 MHz and a pulse width of 0.5 us. Ap- proximately how many osci

podryga [215]

At the frequency of 5 MHz, the period of the oscillations is 1/5meg. That's a period of 1/5 microsecond.

There are 5 full cycles in one full microsecond, and there are 2.5 full cycles in a 0.5 us pulse.

You'll have to decide for yourself how damped a pulse of 2.5 cycles is, because the parameters of the definition are corrupted in the question.

4 0

2 years ago